Answer to Question #126027 in Trigonometry for Habeeb goat

Question

In a triangle, suppose A=37∘, a=3, and b=4.

Find B, C, and c.

SHOW ALL WORK TO RECEIVE FULL CREDIT

If there are no solutions, show why and then write NO SOLUTIONS as your final answer.

If there are two solutions, find B1, C1, and c1

and then find B2, C2, and c2

.

Answer 1

1) According to the Law of cosines:

$a^2=c^2+b^2-2bc\sdot cosA$

$3^2=c^2+4^2-2\sdot4c\sdot cos37^o$

$c^2-8c\sdot0.8+16-9=0$

$c^2-6.4c+7=0$

We got a quadratic equation $ax^2+bx+c=0$

Its common solution will be 

$x_1,_2={-b±\sqrt{b^2-4ab}\over2a}$

In our case its particular solutions are

$c_1={{6.4+\sqrt{6.4^2-4\sdot1\sdot7}}\over{2\sdot1}}={{{6.4+\sqrt{12.96}}\over{2}}={{6.4+3.6}\over{2}}}=3.2+1.8=5$

$c_2={{6.4-\sqrt{6.4^2-4\sdot1\sdot7}}\over{2\sdot1}}={{6.4-\sqrt{12.96}}\over{2}}={{6.4-3.6}\over{2}}=3.2-1.8=1.4$

2) According to the Law of sines:

${sinA\over{a}}={sinB\over{b}}={sinC\over{c}}$

Now we will find B, C

$B_1={arcsin⁡({{bsinA}\over{a}})}={arcsin⁡({{4\sdot{3\over{5}}}\over{3}})}={arcsin⁡({4\over{5}})}=53^o$

$C_1={arcsin⁡({{csinA}\over{a}})}={arcsin⁡({{5\sdot{3\over{5}}}\over{3}})}={arcsin⁡(1)}=90^o$

$B_2={arcsin⁡({{bsinA}\over{a}})}={arcsin⁡({{4\sdot{3\over{5}}}\over{3}})}={arcsin⁡({4\over{5}})}=127^o$

$C_2={arcsin⁡({{bsinA}\over{a}})}={arcsin⁡({{1,4\sdot{3\over{5}}}\over{3}})}={arcsin⁡({7\over{25}})}=16^o$